Miscellaneous · Q7: Solve a system in $\frac1x$⁠, $\frac1y$⁠, $\frac1z$

Basics you need

Solution

1. Substitute and write $AX = B$

   Let $u = \frac1x$⁠, $v = \frac1y$⁠, $w = \frac1z$⁠. The system becomes

   $$\begin{aligned} 2u + 3v + 10w &= 4 \\ 4u - 6v + 5w &= 1 \\ 6u + 9v - 20w &= 2 \end{aligned}$$

   that is, $AX = B$ with

   $$A = \begin{bmatrix} 2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20 \end{bmatrix}, \quad X = \begin{bmatrix} u \\ v \\ w \end{bmatrix}, \quad B = \begin{bmatrix} 4 \\ 1 \\ 2 \end{bmatrix}.$$
   $\frac{2}{x} = 2\cdot\frac1x = 2u$⁠, and similarly for every term, so each equation is linear in $u, v, w$⁠. The orders fit: A is $3\times3$ and X is $3\times1$⁠, so AX is $3\times1$⁠, the order of B. The equations only make sense for $x, y, z \ne 0$⁠; conversely, any solution $(u, v, w)$ with all three non-zero gives back $x = \frac1u$⁠, $y = \frac1v$⁠, $z = \frac1w$⁠. So the two systems have exactly the same solutions, provided we check $u, v, w \ne 0$ at the end.
   Section 4.6.1: the system written as $AX = B$⁠. Lesson: solving systems AX = B

2. The determinant of A

   No entry is zero; expand along $R_1 = (2, 3, 10)$⁠, with signs $+, -, +$⁠:

   $$M_{11} = \begin{vmatrix} -6 & 5 \\ 9 & -20 \end{vmatrix} = 120 - 45 = 75$$ $$M_{12} = \begin{vmatrix} 4 & 5 \\ 6 & -20 \end{vmatrix} = -80 - 30 = -110$$ $$M_{13} = \begin{vmatrix} 4 & -6 \\ 6 & 9 \end{vmatrix} = 36 + 36 = 72$$ $$\begin{aligned} |A| &= 2(75) - 3(-110) + 10(72) \\ &= 150 + 330 + 720 = 1200. \end{aligned}$$

   Since $|A| = 1200 \ne 0$⁠, A is non-singular: the system has a unique solution.

   In $M_{11}$⁠, $(-6)(-20) = 120$ and $5\cdot9 = 45$⁠. In $M_{13}$⁠, $4\cdot9 = 36$ and $(-6)\cdot6 = -36$⁠, and subtracting $-36$ adds 36. A non-zero determinant is Case I of Section 4.6.1: $A^{-1}$ exists and gives the one and only solution.
   Section 4.2.3 (expansion); Section 4.6.1, Case I. Lesson: expanding a 3 × 3 determinant

3. All cofactors, adj A and $A^{-1}$

   Row 1 of minors is in Step 2. The others (rows of A: $(2, 3, 10)$⁠, $(4, -6, 5)$⁠, $(6, 9, -20)$⁠):

   $$\begin{aligned} M_{21} &= \begin{vmatrix} 3 & 10 \\ 9 & -20 \end{vmatrix} = -150 \\ M_{22} &= \begin{vmatrix} 2 & 10 \\ 6 & -20 \end{vmatrix} = -100 \\ M_{23} &= \begin{vmatrix} 2 & 3 \\ 6 & 9 \end{vmatrix} = 0 \\ M_{31} &= \begin{vmatrix} 3 & 10 \\ -6 & 5 \end{vmatrix} = 75 \\ M_{32} &= \begin{vmatrix} 2 & 10 \\ 4 & 5 \end{vmatrix} = -30 \\ M_{33} &= \begin{vmatrix} 2 & 3 \\ 4 & -6 \end{vmatrix} = -24 \end{aligned}$$

   With the signs $(-1)^{i+j}$⁠:

   $$\begin{aligned} A_{11} &= 75,\quad A_{12} = 110,\quad A_{13} = 72 \\ A_{21} &= 150,\quad A_{22} = -100,\quad A_{23} = 0 \\ A_{31} &= 75,\quad A_{32} = 30,\quad A_{33} = -24 \end{aligned}$$

   Transpose the cofactor matrix to get the adjoint, then divide by $|A|$⁠:

   $$\operatorname{adj}A = \begin{bmatrix} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{bmatrix}, \quad A^{-1} = \frac{1}{1200}\operatorname{adj}A.$$
   For example, $M_{21} = 3(-20) - 10\cdot9$ $= -60 - 90 = -150$⁠, so $A_{21} = (-1)^3(-150) = 150$⁠; and $M_{23} = 2\cdot9 - 3\cdot6 = 0$⁠, because rows 1 and 3 of A agree in their first two entries up to the factor 3. Row 1 of adj A is $(A_{11}, A_{21}, A_{31})$⁠: the cofactors of column 1, because of the transpose.
   Section 4.4, Definitions 1, 2; Section 4.5.1, Definition 3; Section 4.5, Theorem 4. Lesson: the adjoint

4. Check $A\,(\operatorname{adj}A) = |A|\,I$

   Rows of A times the columns of adj A, which are $(75, 110, 72)$⁠, $(150, -100, 0)$⁠, $(75, 30, -24)$⁠:

   $$\begin{aligned} (1,1)&: 150 + 330 + 720 = 1200 \\ (1,2)&: 300 - 300 + 0 = 0 \\ (1,3)&: 150 + 90 - 240 = 0 \\ (2,1)&: 300 - 660 + 360 = 0 \\ (2,2)&: 600 + 600 + 0 = 1200 \\ (2,3)&: 300 - 180 - 120 = 0 \\ (3,1)&: 450 + 990 - 1440 = 0 \\ (3,2)&: 900 - 900 + 0 = 0 \\ (3,3)&: 450 + 270 + 480 = 1200 \end{aligned}$$

   So $A\,(\operatorname{adj}A) = 1200\,I$ ✓, which confirms every cofactor and gives $AA^{-1} = I$⁠.

   This is Theorem 1 of Section 4.5. Dividing both sides by 1200 shows that $\frac{1}{1200}\operatorname{adj}A$ really is $A^{-1}$⁠. For instance, entry $(2,1)$ is $4\cdot75 + (-6)\cdot110 + 5\cdot72$ $= 300 - 660 + 360$⁠.
   Section 4.5, Theorem 1. Lesson: inverse of a matrix

5. Solve $X = A^{-1}B$

   First multiply adj A by $B = (4, 1, 2)$⁠:

   $$\begin{aligned} \text{row 1}&: 75\cdot4 + 150\cdot1 + 75\cdot2 = 600 \\ \text{row 2}&: 110\cdot4 - 100\cdot1 + 30\cdot2 = 400 \\ \text{row 3}&: 72\cdot4 + 0\cdot1 - 24\cdot2 = 240 \end{aligned}$$

   Then divide by 1200:

   $$X = \begin{bmatrix} u \\ v \\ w \end{bmatrix} = \frac{1}{1200}\begin{bmatrix} 600 \\ 400 \\ 240 \end{bmatrix} = \begin{bmatrix} \frac12 \\[2pt] \frac13 \\[2pt] \frac15 \end{bmatrix}.$$
   From $AX = B$⁠, multiplying on the left by $A^{-1}$ gives $A^{-1}AX = A^{-1}B$⁠, that is, $X = A^{-1}B$⁠. The inverse is unique, so this is the only solution. Multiplying by adj A first and dividing by 1200 last keeps the arithmetic in whole numbers: $400 \div 1200 = \frac13$ and $240 \div 1200 = \frac15$⁠.
   Section 4.6.1, Case I: $X = A^{-1}B$⁠. Lesson: solving systems AX = B

6. Go back to $x$⁠, $y$⁠, $z$

   All three of $u = \frac12$⁠, $v = \frac13$⁠, $w = \frac15$ are non-zero, so we may take reciprocals:

   $$x = \frac1u = 2, \quad y = \frac1v = 3, \quad z = \frac1w = 5.$$
   We set $u = \frac1x$⁠, so $x = \frac1u$⁠, which needs $u \ne 0$⁠. Had any of $u, v, w$ been 0, there would be no matching $x$⁠, $y$ or $z$ and the original system would have no solution. Here all is well, and since $(u, v, w)$ is unique, so is $(x, y, z)$⁠.
   Reciprocals. Basics: algebra the chapter leans on

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